# Automorphism with no square root

Does it exist an automorphism $$f$$ in a separable $$\mathbb C$$ Hilbert space, such that $$f$$ has no square root?

If so, a concrete example would be useful.

• Certainly yes for a real Hilbert space, for example let $H=\Bbb R$ and define $T:H\to H$ by $Tx=-x$. – David C. Ullrich Nov 8 at 20:49
• I should have mentioned over $\mathbb C$. I edited the question. Thanks David. – user384617 Nov 8 at 21:12
• The shift operators on $l^2$ do not have square roots: math.stackexchange.com/questions/485259/… The proofs there should work for $l^2$ equally well – daw Nov 8 at 21:33
• Thanks but these are not automorphisms – user384617 Nov 8 at 22:20
• The shift operator on $\ell_2(\Bbb Z)$ is an automorphism. Of course it has a square root, and in fact it's normal... – David C. Ullrich Nov 9 at 1:55

Concretely, given a domain $$D\subset\mathbb C$$ define $$D^{1/2}=\{\lambda\in\mathbb C:\ \lambda^2\in D\}.$$ They proved that the multiplication operator $$M_z\in B(L^2(D))$$ given by $$(M_zf)(z)=zf(z)$$ has a square root if and only if $$D^{1/2}$$ is disconnected. This can be seen to be equivalent to $$D$$ surrounding (but not containing, obviously) the origin.
So, if for instance you take any disk that does not contain the origin, say $$D=\{\lambda:\ |\lambda-2|<1\}$$, then $$M_z\in B(L^2(D))$$ is invertible and has no square root.